#include<bits/stdc++.h>
using namespace std;
double ret=-11625907.5798;
//假设0出现的次数为x，则1出现的次数为n-x;
//x<n/2;
int n=23333333;
int main()
{
    //不断枚举0的数量
    int l=0,r=n/2;
    for(int i=l;i<=r;i++)
    {
        double p0=(double)i/(double)n;
        double p1=(double)(n-i)/(double)n;
        double ret1=(double)i*p0*log2(p0)+(double)(n-i)*p1*log2(p1);
        if(abs(ret1-ret)<=1e-4)
        {
            cout<<i<<endl;
            break;
        }
    }
    return 0;
}


